julian403
Full Member level 5
- Joined
- Feb 28, 2014
- Messages
- 254
- Helped
- 1
- Reputation
- 2
- Reaction score
- 1
- Trophy points
- 18
- Location
- Argentina
- Activity points
- 2,105
We know that the mains voltage is 110 sen (2 Π 60 t)
And by the faraday's law:
\[V = - \frac{ d}{ d t} {\int}_{S} \vec{B} . d \vec{S}\]
\[B\] is proportional to current. \[B = \mu \frac{N I}{L}\] where L is the magnetic path.
\[V = - \frac{ d}{ d t} {\int}_{S} \mu \frac{N I}{L}. d \vec{S}\]
And as the magnetic field in the core is constants:
\[V = - \frac{ d}{ d t} \mu \frac{N I}{L} A\]
The only staff which can changes is the current, so:
\[V = - \mu \frac{N }{L} A \frac{ d I}{ d t}\]
So, in the transformer there is:
\[{V}_{1} = - \mu \frac{{N}_{1} }{L} A \frac{ d{I}_{1}}{d t}\]
\[{V}_{2} = - \mu \frac{{N}_{2} }{L} A \frac{ d {I}_{2}}{d t}\]
And here is my question, the core is made of silicon steel which \[{\mu}_{r} \] ≈ 14000 and the saturation magnetici field is 2 [T].
So B can not be bigger than 2[T]
\[110 sen (2 \pi 60 t) = - 2 \pi 60 \mu \frac{{N}_{1} {I}_{1}}{L} A sen (2 \pi 60 t)\]
And \[B =\mu \frac{{N}_{1} {I}_{1}}{L} \] ≦ 2[T]
\[110 sen (2 \pi 60 t) = - 2 \pi 60 (2[T]) A sen (2 \pi 60 t)\]
I that way A must be \[0.14 [{m}^{2}]\] and that's bigger that the comertial transformer. Why?
And by the faraday's law:
\[V = - \frac{ d}{ d t} {\int}_{S} \vec{B} . d \vec{S}\]
\[B\] is proportional to current. \[B = \mu \frac{N I}{L}\] where L is the magnetic path.
\[V = - \frac{ d}{ d t} {\int}_{S} \mu \frac{N I}{L}. d \vec{S}\]
And as the magnetic field in the core is constants:
\[V = - \frac{ d}{ d t} \mu \frac{N I}{L} A\]
The only staff which can changes is the current, so:
\[V = - \mu \frac{N }{L} A \frac{ d I}{ d t}\]
So, in the transformer there is:
\[{V}_{1} = - \mu \frac{{N}_{1} }{L} A \frac{ d{I}_{1}}{d t}\]
\[{V}_{2} = - \mu \frac{{N}_{2} }{L} A \frac{ d {I}_{2}}{d t}\]
And here is my question, the core is made of silicon steel which \[{\mu}_{r} \] ≈ 14000 and the saturation magnetici field is 2 [T].
So B can not be bigger than 2[T]
\[110 sen (2 \pi 60 t) = - 2 \pi 60 \mu \frac{{N}_{1} {I}_{1}}{L} A sen (2 \pi 60 t)\]
And \[B =\mu \frac{{N}_{1} {I}_{1}}{L} \] ≦ 2[T]
\[110 sen (2 \pi 60 t) = - 2 \pi 60 (2[T]) A sen (2 \pi 60 t)\]
I that way A must be \[0.14 [{m}^{2}]\] and that's bigger that the comertial transformer. Why?